Show that if ##x>1##, ##\log_e\sqrt{x^2-1}=\log_ex-\dfrac{1}{2x^2}-##

[RChristenk]

Homework Statement

Show that if $x>1$, $\log_e\sqrt{x^2-1}=\log_ex-\dfrac{1}{2x^2}-\dfrac{1}{4x^2}-\dfrac{1}{6x^6}-\cdots$

Relevant Equations

Logarithm Rules, Logarithmic Series

$\log_e\sqrt{x^2-1}=\dfrac{1}{2}[\log_e[(x+1)(x-1)]]=\dfrac{1}{2}[\log_e(x+1)+\log_e(x-1)]$.

$\Rightarrow \log_e(x-1)=\log_e[x(1-\dfrac{1}{x})]=\log_ex+\log_e(1-\dfrac{1}{x})$

We know:

$\log_e(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots$

$\log_e(1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-\cdots$

Hence:

$\dfrac{1}{2}[\log_e(x+1)+\log_e(x-1)]=\dfrac{1}{2}[\log_e(1+x)+\log_ex+\log_e(1-\dfrac{1}{x})]$

$\Rightarrow \dfrac{1}{2}(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots+\log_ex-\dfrac{1}{x}-\dfrac{(\dfrac{1}{x})^2}{2}-\dfrac{(\dfrac{1}{x})^3}{3}-\cdots)$

$\Rightarrow \dfrac{1}{2}(\log_ex+x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots-\dfrac{1}{x}-\dfrac{1}{2x^2}-\dfrac{1}{3x^3}-\cdots)$

This is as far as I got. I'm not sure how to perform the necessary algebra to get to

$\log_ex-\dfrac{1}{2x^2}-\dfrac{1}{4x^2}-\dfrac{1}{6x^6}-\cdots$.

 

[anuttarasammyak]

1/x^2 and higher orders of this are small for large positive x. The series is difference of LHS from log x . We know that LHS is a little smaller than log x.

 

[PeroK]

Homework Statement: Show that if $x>1$, $\log_e\sqrt{x^2-1}=\log_ex-\dfrac{1}{2x^2}-\dfrac{1}{4x^2}-\dfrac{1}{6x^6}-\cdots$
Relevant Equations: Logarithm Rules, Logarithmic Series

$\log_e\sqrt{x^2-1}=\dfrac{1}{2}[\log_e[(x+1)(x-1)]]=\dfrac{1}{2}[\log_e(x+1)+\log_e(x-1)]$.

This is good until here. Note, however, that the expansion for $\log_e(1 + x)$ is only valid for $x < 1$. You need to take a factor of $x$ out of both these logs.

 

[pasmith]

For large $x$, $\ln\sqrt{x^2 - 1} \approx \ln x$ so start with $$\begin{split} \ln \sqrt{x^2 - 1} &= \ln\left(x \sqrt{1 - \frac1{x^2}}\right) \\ &=\ln x + \frac12 \ln\left(1 - \frac 1{x^2}\right)\end{split}$$ and since $1/x^2 < 1$ you can expand $\ln\left(1 - \frac 1{x^2}\right)$ without further manipulation.

 

[RChristenk]

Thanks for your replies. Why is $\log_e(1+x)$ invalid for $x \geq 1$?

 

[PeroK]

Thanks for your replies. Why is $\log_e(1+x)$ invalid for $x \geq 1$?

The power series expansion is invalid for $x > 1$, because it has a radius of convergence of $1$.

 

[Mark44]

Thread moved.
Problems involving Taylor or Maclaurin series entail knowledge of calculus.

... $\log_e(1+x)$ ...

@RChristenk, why do you write this as $\log_e(\dots)$ instead of $\ln(\dots)$?

 

[RChristenk]

I'm looking at this free book online:

https://archive.org/details/elementaryalgebr00hall

 

[pasmith]

I'm looking at this free book online:

https://archive.org/details/elementaryalgebr00hall

This text was published in 1896, and use of now outdated notation is not its worst shortcoming by modern standards. The notation $\ln$ is nowhere mentioned; $\log$ itself is reserved for the base 10 logarithm.

Relevant to this problem, which is example 6 on page 434, in relation to the series for $\ln (1 + x)$ the book states at para 540 on p. 433 that "except when $x$ is vey small, [this series] is of little use for numerical calculations" and then proceeds to construct the series for $\ln (N + 1) - \ln N$ which was examined in another thread as a method to find a series for $\ln 2$ (example 3 on page 434).

However the book does not state that the radius of convergence is 1. Indeed the chapter on convergence and divergence of series does not expressly discuss power series or radii of convergence as such, although the examples of the use of the ratio test[* do find radii of convergence of power series - see, eg. p. 382. So a careful reader could possibly determine for themselves that the log series does not converge for $|x| > 1$, although there is no exercise instructing them to do so. The chapter ends with a note referring readers wanting further information on the subject to chapter XXI of the same authors' Higher Algebra.

I would suggest studying from a more modern text, which uses now-standard terminology and notation.

*The test is not expressly named as such; all of the convergence tests are named "the Nth test" in the order in which they are presented.